What is the Value of Cos 45 Degrees ?

Solution :

The value of cos 45 degrees is \(1\over \sqrt{2}\).

Proof :

Let ABC be a triangle, right angled at B, in which \(\angle\) A = \(\angle\) C = 45 degrees45 degrees

\(\therefore\)  BC = AB

Let  AB = BC = a

Then by pythagoras theorem,

\(AC^2\) = \(AB^2\) + \(BC^2\) = \(a^2\) + \(a^2\) = \(2a^2\)

\(\implies\)  AC = \(\sqrt{2}a\)

In \(\Delta\) ABC,  \(\angle\) C = 45 degrees

By using trigonometric formulas,

\(cos 45^{\circ}\) = \(base\over hypotenuse\) = \(b\over h\)

\(cos 45^{\circ}\)  = side adjacent to 45 degrees/hypotenuse = \(BC\over AC\) = \(a\over \sqrt{2}a\) = \(1\over \sqrt{2}\)

Hence, the value of \(cos 45^{\circ}\) = \(1\over \sqrt{2}\)

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